The rate constant for the reaction:
`2N_(2)O_(5) rarr 4NO_(2)+O_(2)` is `3.0xx10^(-5) sec^(-1)`. If the rate is `2.40xx10^(-5) M sec^(-1)`, then the concentration of `N_(2)O_(5)` (in M) is:`

Options

A. (a) `1.4`
B. (b) `1.2`
C. (c ) `0.04`
D. (d) `0.8`

Answer

d

Explanation

`r=K[N_(2)O_(5)]` for `I` order as unit of `K= sec^(-1)`
`2.40xx10^(-5)=3.0xx10^(-5) [N_(2)O_(5)]`
`:. [N_(2)O_(5)]=2.40/3.0=0.8M`